Problem 52 - Applied Calculus

Question

Evaluate each definite integral. $$ \int_{1}^{4} \frac{1}{\sqrt{z}} d z $$

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Identify the Integral

The integral we are asked to evaluate is \( \int_{1}^{4} \frac{1}{\sqrt{z}} \, dz \). This is a definite integral with the function \( \frac{1}{\sqrt{z}} \) as the integrand and the limits of integration from 1 to 4.

Step 2: Rewrite the Integrand

Rewrite \( \frac{1}{\sqrt{z}} \) as \( z^{-1/2} \) to make it easier to integrate. So, the integral becomes \( \int_{1}^{4} z^{-1/2} \, dz \).

Step 3: Integrate the Function

Use the power rule for integration: \( \int z^n \, dz = \frac{z^{n+1}}{n+1} + C \). For \( z^{-1/2} \), \( n = -1/2 \), so the integral becomes \( \frac{z^{1/2}}{1/2} = 2z^{1/2} \).

Step 4: Evaluate the Definite Integral

Substitute the limits of integration into the evaluated antiderivative \( 2z^{1/2} \). Calculate \( 2(4^{1/2}) - 2(1^{1/2}) \).

Step 5: Calculate with Limits

Perform the calculations: \( 2 \times 2 - 2 \times 1 = 4 - 2 \). This evaluates to 2.

Key Concepts

Integration by Parts

Integration by Parts is a technique used to integrate products of functions. It's especially useful when dealing with integrals that are products of an algebraic function and a transcendental function, such as logarithmic, exponential, or trigonometric functions. The formula for Integration by Parts comes from the product rule for differentiation and is expressed as:\[\int u \, dv = uv - \int v \, du\]Here:

• \( u \) is a function you choose, often one that becomes simpler when differentiated.
• \( dv \) is the remaining part of the integrand.
• \( du \) is the derivative of \( u \).
• \( v \) is the antiderivative of \( dv \).

Applying integration by parts can sometimes require multiple steps or iterations if the resulting integral is still complex. However, for simpler forms, choosing the right \( u \) and \( dv \) can simplify the process significantly.
In our original exercise, we didn't need integration by parts because it wasn't required to evaluate a simple power function. This technique shines in more complex scenarios, illustrating the beauty and depth of calculus problem-solving strategies.

Power Rule for Integration

The Power Rule for Integration is a fundamental and straightforward method to find the integral of power functions. It is particularly useful when integrating functions of the form \( z^n \). The rule states:\[ \int z^n \, dz = \frac{z^{n+1}}{n+1} + C \]where:

• \( n eq -1 \) to avoid dividing by zero.
• \( C \) is the constant of integration for indefinite integrals.

In the given problem, we apply the Power Rule to the integrand \( z^{-1/2} \). Identifying \( n = -1/2 \), we find its integral becomes \( \frac{z^{1/2}}{1/2} \), which simplifies to \( 2z^{1/2} \).
This process transforms our problem into calculating limits, which brings us to the next aspect of definite integration. This foundational technique wonderfully illustrates how even complex-looking integrals can have elegantly simple solutions with the power rule.

Limits of Integration

Limits of integration, in the context of definite integrals, define the interval over which the integration is performed. These limits are represented as the lower and upper bounds \( [a, b] \). Evaluating a definite integral involves calculating the antiderivative of the integrand at these bounds and taking their difference:\[ \int_{a}^{b} f(z) \, dz = F(b) - F(a) \]where:

• \( F(z) \) is the antiderivative of \( f(z) \).
• \( F(b) \) and \( F(a) \) are the values of the antiderivative at the upper and lower limits, respectively.

In our exercise, after finding the antiderivative as \( 2z^{1/2} \), we substitute the limits \( 4 \) and \( 1 \) into the antiderivative to find \( 2(4^{1/2}) - 2(1^{1/2}) \). Performing these calculations gives us the result \( 4 - 2 = 2 \).
The technique emphasizes how specific calculations connected with integration can simplify what seems complex at first glance by merely substituting and then evaluating simple arithmetic operations, bringing the topic full circle and allowing us to appreciate calculus' efficiency.